Let $G$ be a finite simple group of Lie type over a finite field of characteristic $p\neq 2$ and $q$ elements. We know that if $p$ is not bad for $G$ then $G$ contains $q^l$ regular unipotent elements where $l$ is the semisimple rank of $G$ and all of them are conjugate in $G$ (See p.130 and p.131 of Carter's book). So $|C_G(u)|=q^l$ where $u$ is a regular unipotent element.
Now let $u_0$ be a $p$-element which its centralizer is a $p$-group. Is it true that $u_0$ is a regular unipotent element?
Distinguished unipotent elements:
I think the answer is close to no: on page 173 Carter describes the $p$-elements so that $C_G(u)_o$ is a $p$-group, the so-called distinguished unipotent elements. Most simple algebraic groups contain more than one conjugacy class of distinguished unipotent elements as on page 174-177.
I checked the results in magma, but things did not work out quite as desired. I haven't found any “really distinguished” classes, where the entire centralizer (not just its connected component) is a $p$-group. [ I've only checked a few cases, Cn and G2; An doesn't work. ]
Simple versus adjoint:
Unfortunately, you have a slight misinterpretation of Carter's statements. In groups like $G=\operatorname{PGL}$, there will be a single conjugacy class of regular unipotent elements, and its centralizer will be a $p$-group. The good news is that the unipotent elements is necessarily contained within the simple group $O^{p'}(G)$ (like $\operatorname{PSL}$). The bad news is that since the centralizer is also a $p$-group, the centralizer is also contained within $O^{p'}(G)$, and so necessarily will split into $[G:O^{p'}(G)]$ classes (such as $\gcd(n,q-1)$ for our running example).
As another example, $\operatorname{PSp}(2n,p^f)$ has two conjugacy classes of elements whose centralizers are of order $p^{2f}=q^2$. I believe it has no other conjugacy classes whose centralizers are $p$-groups (it does not for small $n,p,f$).